I want regex to match web addresses such as http://www.example.com, example.co.uk, en.example.com etc. I've been using ^(https?://|www\.|)[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(/\S*)?$ and testing it on http://regexpal.com/, and it seems to work exactly as it should.
However, when I put it in autohotkey, it seems to match extra things like example and example.something, when it shouldn't. It then doesn't match things like example.com/something and example.com/something.html when it should.
If RegExMatch(Clipboard, "^(https?://|www\.|)[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(/\S*)?$")
Msgbox, it matches
else
Msgbox, it doesn't
Matching URLs, host names etc is a problem solved many times; I suggest you adapt some standard regex. Perhaps SO question: Fully qualified domain name validation is helpful.
If you're composing the regex as an exercise:
Does it really match the string example? You firmly assert the string to contain a ., so it never should. Maybe AHK doesn't escape . the standard way?
If [a-zA-Z]{2,3} should match top level domain, you forgot about .info.
You may want to allow strings of whitespace of arbitrary length at the end and beginning, if you accidentally copied some such into the clipboard. I.e. ^\s*your-regex-thingy\s*$
example.something is a match, because it begins with the empty string, follows with a sequence of 1 or more alphanumerics (or -, .), one ., 2 or 3 letters, and ends with a sequence of non-whitespace.
example.com/something.html might fail to match if the entire substring example.com is matched by the group [a-zA-Z0-9\-\.]+. It shouldn't if the regex engine is correctly implemented, though. Perhaps you need to escape +, | or some such, engines have varying conventions on such (i.e. sed and pcre have differing opinions on + and ( if I'm not mistaken.
Related
I am trying to regex the following string:
https://www.amazon.com/Tapps-Top-Apps-and-Games/dp/B00VU2BZRO/ref=sr_1_3?ie=UTF8&qid=1527813329&sr=8-3&keywords=poop
I want only B00VU2BZRO.
This substring is always going to be a 10 characters, alphanumeric, preceded by dp/.
So far I have the following regex:
[d][p][\/][0-9B][0-9A-Z]{9}
This matches dp/B00VU2BZRO
I want to match only B00VU2BZRO with no dp/
How do I regex this?
Here is one regex option which would produce an exact match of what you want:
(?<=dp\/)(.*)(?=\/)
Demo
Note that this solution makes no assumptions about the length of the path fragment occurring after dp/. If you want to match a certain number of characters, replace (.*) with (.{10}), for example.
Depending on your language/method of application, you have a couple of options.
Positive look behind. This will make your regex more complicated, but will make it match what you want exactly:
(<=dp/)[0-9A-Z]{10}
The construct (<=...) is called a positive look behind. It will not consume any of the string, but will only allow the match to happen if the pattern between the parens is matched.
Capture group. This will make the regex itself slightly simpler, but will add a step to the extraction process:
dp/([0-9A-Z]{10})
Anything between plain parens is a capture group. The entire pattern will be matched, including dp/, but most languages will give you a way of extracting the portion you are interested in.
Depending on your language, you may need to escape the forward slash (/).
As an aside, you never need to create a character class for single characters: [d][p][\/] can equally well be written as just dp\/.
My string being of the form:
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
I only want to match against the last segment of whitespace before the last period(.)
So far I am able to capture whitespace but not the very last occurrence using:
\s+(?=\.\w)
How can I make it less greedy?
In a general case, you can match the last occurrence of any pattern using the following scheme:
pattern(?![\s\S]*pattern)
(?s)pattern(?!.*pattern)
pattern(?!(?s:.*)pattern)
where [\s\S]* matches any zero or more chars as many as possible. (?s) and (?s:.) can be used with regex engines that support these constructs so as to use . to match any chars.
In this case, rather than \s+(?![\s\S]*\s), you may use
\s+(?!\S*\s)
See the regex demo. Note the \s and \S are inverse classes, thus, it makes no sense using [\s\S]* here, \S* is enough.
Details:
\s+ - one or more whitespace chars
(?!\S*\s) - that are not immediately followed with any 0 or more non-whitespace chars and then a whitespace.
You can try like so:
(\s+)(?=\.[^.]+$)
(?=\.[^.]+$) Positive look ahead for a dot and characters except dot at the end of line.
Demo:
https://regex101.com/r/k9VwC6/3
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=((?<=\S)\s+)).*
replaced by `>\1<`
> <
As a more generalized example
This example defines several needles and finds the last occurrence of either one of them. In this example the needles are:
defined word findMyLastOccurrence
whitespaces (?<=\S)\s+
dots (?<=[^\.])\.+
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)).*
replaced by `>\1<`
>..<
Explanation:
Part 1 .*
is greedy and finds everything as long as the needles are found. Thus, it also captures all needle occurrences until the very last needle.
edit to add:
in case we are interested in the first hit, we can prevent the greediness by writing .*?
Part 2 (?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+|(?<=**Not**NeedlePart)NeedlePart+))
defines the 'break' condition for the greedy 'find-all'. It consists of several parts:
(?=(needles))
positive lookahead: ensure that previously found everything is followed by the needles
findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)|(?<=**Not**NeedlePart)NeedlePart+
several needles for which we are looking. Needles are patterns themselves.
In case we look for a collection of whitespaces, dots or other needleparts, the pattern we are looking for is actually: anything which is not a needlepart, followed by one or more needleparts (thus needlepart is +). See the example for whitespaces \s negated with \S, actual dot . negated with [^.]
Part 3 .*
as we aren't interested in the remainder, we capture it and dont use it any further. We could capture it with parenthesis and use it as another group, but that's out of scope here
SIMPLE SOLUTION for a COMMON PROBLEM
All of the answers that I have read through are way off topic, overly complicated, or just simply incorrect. This question is a common problem that regex offers a simple solution for.
Breaking Down the General Problem
THE STRING
The generalized problem is such that there is a string that contains several characters.
THE SUB-STRING
Within the string is a sub-string made up of a few characters. Often times this is a file extension (i.e .c, .ts, or .json), or a top level domain (i.e. .com, .org, or .io), but it could be something as arbitrary as MC Donald's Mulan Szechuan Sauce. The point it is, it may not always be something simple.
THE BEFORE VARIANCE (Most important part)
The before variance is an arbitrary character, or characters, that always comes just before the sub-string. In this question, the before variance is an unknown amount of white-space. Its a variance because the amount of white-space that needs to be match against varies (or has a dynamic quantity).
Describing the Solution in Reference to the Problem
(Solution Part 1)
Often times when working with regular expressions its necessary to work in reverse.
We will start at the end of the problem described above, and work backwards, henceforth; we are going to start at the The Before Variance (or #3)
So, as mentioned above, The Before Variance is an unknown amount of white-space. We know that it includes white-space, but we don't know how much, so we will use the meta sequence for Any Whitespce with the one or more quantifier.
The Meta Sequence for "Any Whitespace" is \s.
The "One or More" quantifier is +
so we will start with...
NOTE: In ECMAS Regex the / characters are like quotes around a string.
const regex = /\s+/g
I also included the g to tell the engine to set the global flag to true. I won't explain flags, for the sake of brevity, but if you don't know what the global flag does, you should DuckDuckGo it.
(Solution Part 2)
Remember, we are working in reverse, so the next part to focus on is the Sub-string. In this question it is .com, but the author may want it to match against a value with variance, rather than just the static string of characters .com, therefore I will talk about that more below, but to stay focused, we will work with .com for now.
It's necessary that we use a concept here that's called ZERO LENGTH ASSERTION. We need a "zero-length assertion" because we have a sub-string that is significant, but is not what we want to match against. "Zero-length assertions" allow us to move the point in the string where the regular expression engine is looking at, without having to match any characters to get there.
The Zero-Length Assertion that we are going to use is called LOOK AHEAD, and its syntax is as follows.
Look-ahead Syntax: (?=Your-SubStr-Here)
We are going to use the look ahead to match against a variance that comes before the pattern assigned to the look-ahead, which will be our sub-string. The result looks like this:
const regex = /\s+(?=\.com)/gi
I added the insensitive flag to tell the engine to not be concerned with the case of the letter, in other words; the regular expression /\s+(?=\.cOM)/gi
is the same as /\s+(?=\.Com)/gi, and both are the same as: /\s+(?=\.com)/gi &/or /\s+(?=.COM)/gi. Everyone of the "Just Listed" regular expressions are equivalent so long as the i flag is set.
That's it! The link HERE (REGEX101) will take you to an example where you can play with the regular expression if you like.
I mentioned above working with a sub-string that has more variance than .com.
You could use (\s*)(?=\.\w{3,}) for instance.
The problem with this regex, is even though it matches .txt, .org, .json, and .unclepetespurplebeet, the regex isn't safe. When using the question's string of...
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
as an example, you can see at the LINK HERE (Regex101) there are 3 lines in the string. Those lines represent areas where the sub-string's lookahead's assertion returned true. Each time the assertion was true, a possibility for an incorrect final match was created. Though, only one match was returned in the end, and it was the correct match, when implemented in a program, or website, that's running in production, you can pretty much guarantee that the regex is not only going to fail, but its going to fail horribly and you will come to hate it.
You can try this. It will capture the last white space segment - in the first capture group.
(\s+)\.[^\.]*$
I am trying to find the appropriate regex pattern that allows me to pick out whole words either starting with or ending with a comma, but leave out numbers. I've come up with ([\w]+,) which matches the first word followed by a comma, so in something like:
red,1,yellow,4
red, will match, but I am trying to find a solution that will match like like the following:
red, 1 ,yellow, 4
I haven't been able to find anything that can break strings up like this, but hopefully you'll be able to help!
This regex
,?[a-zA-Z][a-zA-Z0-9]*,?
Matches 'words' optionally enclose with commas. No spaces between commas and the 'word' are permitted and the word must start with an alphanumeric.
See here for a demo.
To ascertain that at least one comma is matched, use the alternation syntax:
(,[a-zA-Z][a-zA-Z0-9]*|[a-zA-Z][a-zA-Z0-9]*,)
Unfortunately no regex engine that i am aware of supports cascaded matching. However, since you usually operate with regexen in the context of programming environments, you could repeatedly match against a regex and take the matched substring for further matches. This can be achieved by chaining or iterated function calls using speical delimiter chars (which must be guaranteed not to occur in the test strings).
Example (Javascript):
"red, 1 ,yellow, 4, red1, 1yellow yellow"
.replace(/(,?[a-zA-Z][a-zA-Z0-9]*,?)/g, "<$1>")
.replace(/<[^,>]+>/g, "")
.replace(/>[^>]+(<|$)/g, "> $1")
.replace(/^[^<]+</g, "<")
In this example, the (simple) regex is tested for first. The call returns a sequence of preliminary matches delimted by angle brackets. Matches that do not contain the required substring (, in this case) are eliminated, as is all intervening material.
This technique might produce code that is easier to maintain than a complicated regex.
However, as a rule of thumb, if your regex gets too complicated to be easily maintained, a good guess is that it hasn't been the right tool in the first place (Many engines provide the x matching modifier that allows you to intersperse whitespace - namely line breaks and spaces - and comments at will).
The issue with your expression is that:
- \w resolves to this: [a-zA-Z0-9_]. This includes numeric data which you do not want.
- You have the comma at the end, this will match foo, but not ,foo.
To fix this, you can do something like so: (,\s*[a-z]+)|([a-z]+\s*,). An example is available here.
I am very new to regex and regular expressions, and I am stuck in a situation where I want to apply a regex on an JSF input field.
Where
alphanumeric
multiple spaces
multiple dot(.)
multiple hyphen (‐)
are allowed, and Minimum limit is 1 and Maximum limit is 5.
And for multiple values - they must be separated by comma (,)
So a Single value can be:
3kd-R
or
k3
or
-4
And multiple values (must be comma separated):
kdk30,3.K-4,ER--U,2,.I3,
By the help of stackoverflow, so far I am able to achieve only this:
(^[a-zA-Z0-9 ]{5}(,[a-zA-Z0-9 ]{5})*$)
Something like
^[-.a-zA-Z0-9 ]{1,5}(,[-.a-zA-Z0-9 ]{1,5})*$
Changes made
[-.a-zA-Z0-9 ] Added - and . to the character class so that those are matched as well.
{1,5} Quantifier, ensures that it is matched minimum 1 and maximum 5 characters
Regex demo
You've done pretty good. You need to add hyphen and dot to that first character class. Note: With the hyphen, since it delegates ranges within a character class, you need to position it where contextually it cannot be specifying a range--not to say put it where it seems like it would be an invalid range, e.g., 7-., but positionally cannot be a range, i.e., first or last. So your first character class would look something like this:
[a-zA-Z 0-9.-]{1,5} or [-a-zA-Z0-9 .]{1,5}
So, we've just defined what one segment looks like. That pattern can reoccur zero or more times. Of course, there are many ways to do that, but I would favor a regex subroutine because this allows code reuse. Now if the specs change or you're testing and realize you have to tweak that segment pattern, you only need to change it in one place.
Subroutines are not supported in BRE or ERE, but most widely-used modern regex engines support them (Perl, PCRE, Ruby, Delphi, R, PHP). They are very simple to use and understand. Basically, you just need to be able to refer to it (sound familiar? refer-back? back-reference?), so this means we need to capture the regex we wish to repeat. Then it's as simple as referring back to it, but instead of \1 which refers to the captured value (data), we want to refer to it as (?1), the capturing expression. In doing so, we've logically defined a subroutine:
([a-zA-Z 0-9.-]{1,5})(,(?1))*
So, the first group basically defines our subroutine and the second group consists of a comma followed by the same segment-definition expression we used for the first group, and that is optional ('*' is the zero-or-more quantifier).
If you operate on large quantities of data where efficiency is a consideration, don't capture when you don't have to. If your sole purpose for using parenthesis is to alternate (e.g., \b[bB](asset|eagle)\b hound) or to quantify, as in our second group, use the (?: ... ) notation, which signifies to the regex engine that this is a non-capturing group. Without going into great detail, there is a lot of overhead in maintaining the match locations--not that it's complex, per se, just potentially highly repetitive. Regex engines will match, store the information, then when the match fails, they "give up" the match and try again starting with the next matching substring. Each time they match your capture group, they're storing that information again. Okay, I'm off the soapbox now. :-)
So, we're almost there. I say "almost" because I don't have all the information. But if this should be the sole occupant of the "subject" (line, field, etc.--the data sample you're evaluating), you should anchor it to "assert" that requirement. The caret '^' is beginning of subject, and the dollar '$' is end of subject, so by encapsulating our expression in ^ ... $ we are asserting that the subject matches in it's entirety, front-to-back. These assertions have zero-length; they consume no data, only assert a relative position. You can operate on them, e.g., s/^/ / would indent your entire document two spaces. You haven't really substituted the beginning of line with two spaces, but you're able to operate on that imaginary, zero-length location. (Do some research on zero-length assertions [aka zero-width assertions, or look-arounds] to uncover a powerful feature of modern regex. For example, in the previous regex if I wanted to make sure I did not insert two spaces on blank lines: s/^(?!$)/ /)
Also, you didn't say if you need to capture the results to do something with it. My impression was it's validation only, so that's not necessary. However, if it is needed, you can wrap the entire expression in capturing parenthesis: ^( ... )$.
I'm going to provide a final solution that does not assume you need to capture but does assume the entire subject should consist of this value:
^([a-zA-Z 0-9. -]{1,5})(?:,(?1))*$
I know I went on a bit, but you said you were new to regex, so wanted to provide some detail. I hope it wasn't too much detail.
By the way, an excellent resource with tutorials is regular-expressions dot info, and a wonderful regex development and testing tool is regex101 dot com. And I can never say enough about stack overflow!
I have cookies in my HTTP header like so:
Set-Cookie: frontend=ovsu0p8khivgvp29samlago1q0; adminhtml=6df3s767g199d7mmk49dgni4t7; external_no_cache=1; ZDEDebuggerPresent=php,phtml,php3
and I need to extract the 26 character string that comes after frontend (e.g. ovsu0p8khivgvp29samlago1q0). The following regular expression matches that for me:
(?<=frontend=)(.*)(?=;)
However, I am using Varnish Cache and can only use a regex replace. Therefore, to extract that cookie value (26 character frontend string) I need to match all characters that do not match that pattern (so I can replace them with '').
I've done a fair bit of Googling but so far have drawn a blank. I've tried the following
Match characters that do not match the pattern I want: [^((?<=frontend=)[A-Za-z0-9]{26}(?=;))] which matches random characters, including the ones I want to preserve
I'd be grateful if someone could point me in the right direction, or note where I might have gone wrong.
The Set-Cookie response header is a bit magical in Varnish, since the backends tend to send multiple headers with the same name. This is prohibited by the RFC, but the defacto way to do it.
If you are using Varnish 3.0 you can use the Header VMOD, it can parse the response and extract what you need:
https://github.com/varnish/libvmod-header
Use regex pattern
^Set-Cookie:.*?\bfrontend=([^;]*)
and the "26 character string that comes after frontend" will be in group 1 (usually referred to in the replacement string as $1)
Do you have control over the replacement string? If so, you can go with Ωmega's answer, and use $1 in your replacement string to write the frontend value back.
Otherwise, you could use this:
^Set-Cookie:.*(?!frontend=)|(?<=frontend=.{26}).*$
This will match everything from the start of the string, until frontend= is encountered. Or it will match everything that has frontend= exactly 26 characters to the left of it and up until the end of the string. If those 26 characters are a variable length, it would get signigicantly more complicated, because only .NET supports variable-length lookbehinds.
For your last question. Let's have a look at your regex:
[^((?<=frontend=)[A-Za-z0-9]{26}(?=;))]
Well, firstly the negative character class [^...] you tried to surround you pattern with, doesn't really work like this. It is still a character class, so it matches only a single character that is not inside that class. But it gets even more complicated (and I wonder why it matches at all). So firstly the character class should be closed by the first ]. This character class matches anything that is not (, ?, <, =, ), a letter or a digit. Then the {26} is applied to that, so we are trying to find 26 of those characters. Then the (?=;) which asserts that those 26 characters are followed by ;. Now comes what should not work. The closing ) should actually throw and error. And the final ] would just be interpreted as a literal ].
There are some regex flavors which allow for nesting of character classes (Java does). In this case, you would simply have a character class equivalent to [^a-zA-Z0-9(){}?<=;]. But as far as I could google it, Varnish uses PCRE, and in PCRE your regex should simply not compile.